Chapter 7. Applications of Trigonometry and Vectors. Section 7.1: Oblique Triangles and the Law of Sines Connections (page 307)

Transcription

1 Chapter 7 Applications of Trigonometry and Vectors Section 7.1: Oblique Triangles and the Law of Sines Connections (page 307) ( a h) x ( a h) ycos θ X =, Y = f secθ ysinθ f secθ ysinθ 1. house: X H ft ; Y H ft forest fire: X 77.5 ft ; Y 596. ft ft Exercises 1. C. C, D C = 95 ; b 13 m; a 11 m F 6. A = 99 ; b 34 cm; c 1 cm 7. B = 37.3 ; a 38.5 ft; b 51.0 ft 8. A = 37. ; a 178 m ; c 44 m 9. C = ; b ft ; c ft 10. A = ; a 8.10 m ; b m 11. B = 18.5 ; a 39 yd; c 30 yd 1. A = ; b 16.1 cm; c 5.8 cm 13. A = ; AB 361 ft ; BC 308 ft 14. C = 91.9 ; BC 490 ft; AB 847 ft 15. B ; a 7.01 m; c 1.37 m F 1. With three sides we have, } } } a b c = =. This does not provide sin A sin C enough information to solve the triangle. Whenever you choose two out of the three ratios to create a proportion, you are missing two pieces of information.. The triangle is not a right triangle. 3. This is not a valid statement. It is true that if you have the measures of two angles, the third can be found. However, if you do not have at least one side, the triangle cannot be uniquely determined. If we consider the congruence axiom involving two angles, ASA, an included side must be considered. 4. No m yd km mi in. 30. In triangle MTL, In right triangle MTR, 93.4 m First location: 5.1 mi Second location: 7. mi mi 16. A = ; b 1.94 cm; c.77 cm 17. A = 34.7 ; a 336 ft ; c 5704 ft 18. C = ; a yd ; b 37 yd 19. C = ; b 83. m; c 415. m 0. B = ; b.04 mm; c mm 35. In either case the distance is approximately 419,000 km compared to the actual value of 406,000 km. 36. approximately 10,85 ft ft ft sq unit 99

2 100 Applications of Trigonometry and Vectors sq unit sq unit m 73 ft 356 cm sq unit B1 = 49.1 ; B = C 1 = 101. ; 14. A 1 = 7.. ; C 1 = 59.6 A = 107.8, ; C = B 1 = 6 30 ; A = o km 7.9 in m cm m 100 m 16. A = 37º50, C= 93º0 17. no such triangle exists. 18. no such triangle exists. 19. B 7.19 and C A = and C = B 1 = 0.6 ; C = ; c 0.6 ft. A 1 = 5.5 ; B = 10. ; b 73.9 yd m 53. a = sin A, b =, and c = sin C. 54. Answers will vary. 55. d sinα sin β sin β α = x ( ) Section 7.: The Ambiguous Case of the Law of Sines 1. A. D 3. (a) 4 < h < 5. (b) h = 4 or h > 5 (c) h < 4 4. (a) none (b) h 5 (c) h 5 3. no such triangle exists. 4. no such triangle exists. 5. B ' ; C 1 = 9 00' c km ; c.88 km 6. A ' ; B 1 = 94 50' b m ; b 4.51 m 7. B ; C = ; c ft 8. B ; A = ; a 98.5 yd 9. A 1 = 53.3 ; C 1 = c m ; A = o C ; c = m 30. C 1 = ; B 1 = ; b cm ; B = 8.51 ; b 1369 cm 31. 1; 90 ; a right triangle Answers will vary m ft ft ; 74 mi

3 Section 7.3: The Law of Cosines 101 a+ b sin A+ 39. Prove that =. b Start with the law of sines. a b A = a = sin A Substitute for a in the expression a + b. b A A + b + b a+ b sin sin = B = B b b b A+ B b sin A+ = = B sin A+ = ( ) a b sin A sin B 40. Prove that =. a+ b sin A+ Start with the law of sines. a b A = a = sin A Substitute for a in the expression a b a+ b. A A b b a b sin sin = B = B a+ b A A + b + b A B b sin A = = A+ b b sin A+ sin A = sin A+ Section 7.3: The Law of Cosines Connections (page 34) 1. Answers will vary.. Answers will vary. Exercises 1. (a) SAS (b) law of cosines. A, C, and c (a) SAA (b) law of sines 3. (a) SSA (b) law of sines 4. (a) SSS (b) law of cosines ( ) ( ) 5. (a) ASA (b) law of sines 6. (a) SSA (b) law of sines 7. (a) ASA (b) law of sines 8. (a) SAS (b) law of cosines a 7.0 ; C 1.4 ; B = a 5.4 ; B 40.7 ; C = B 53.1 ; C = 53.1 ; A = 73.7 The angles may not sum to 180 due to rounding. 16. B 108. ; A.3 ; C = b ; A 56.7 ; C = C 95.7 ; A 33.6 ; B = a.60 yd ; B 45.1 ; C = c.83 in. ; A 44.9 ; B = c 6.46 m ; A 53.1 ; B = a 43.7 km ; B 53. ; C = A 8 ; B 37 ; C = C 98 ; A 9 ; B = C ; B ; A = C ; B ; A = C ; B ; A = B ; C ; A = a 156 cm ; C = ; B = c 348 ft ; B = ; A = b 9.59 in. ; C ; A = c 4.76 mi ; A ; B = a 15.7 m ; B 1.6 ; C = b 34.1 cm ; A 5. ; C = C 94 ; A 30 ; B = C 15 ; A 4 ; B = 31

4 10 Applications of Trigonometry and Vectors 37. The value of cos θ will be greater than 1; your calculator will give you an error message (or a nonreal, complex number) when using the inverse cosine function. 38. Answers will vary m cm and 8.8 cm km mi miles ft. 45. θ AB 18 ft 47. A 36 ; B about 5500 meters long 49. second base is 66.8 ft and the distance to both first and third base is 63.7 ft feet. 51. v 39. km m apart approximately 180 mi feet mi 57. A km 59. ft m 61. θ θ sq units ft (rounded to three significant digits) in mi Since the perimeter and area both equal 36 feet, the triangle is a perfect triangle. 74. (a) A = 66, which is an integer (b) (c) (d) ,000 mi cans A = 84, which is an integer A = 4, which is an integer A = 36, which is an integer 77. (a) C 1 = ; B 9. both appear possible. (b) 9. (c) With the law of cosines, we are required to find the inverse cosine of a negative number; therefore; we know angle C is greater than Using the law of cosines, we have a + c b cos B = ac cos B = = = ( )( ) b + c a cos A = bc cos A = = = ( )( ) Since cos B 1= 1= = = = cos A, A is twice the size of B sq units m (rounded to two significant digits) 310 in. (rounded to two significant digits) 1,600 cm 8 yd

5 Section 7.4: Vectors, Operators, and the Dot Product Given point D is on side AB of triangle ABC such that CD bisects C, AD b m ACD = m DCB. Show that =. DB a Let θ = m ACD, and α = m ADC Then θ = m DCB and = m BDC = 180 α. By the law of sines, we have sinθ sinα AD sinα = sinθ = and AD b b sinθ sin ( 180 α) = DB a DBsin ( 180 α) sinθ = a By substitution, we have AD sinα DB sin ( 180 α) = b a AD DB Since sinα = sin ( 180 α), =. b a b Multiplying both sides by, we have DB AD b DB b AD b = =. b DB a DB DB a 80. Each side of the equation leads to the value Chapter 7: Quiz (Sections ) 1. A = a 01 m 3. C sq units 5. A = 189 km 6. A 41.6 or A C = 8 a 648 b mi Section 7.4: Vectors, Operations, and the Dot Product 1. m and p; n and r.. m and q, p and q, n and s, r and s. 3. m = 1; p m = ; t n = 1; r p = t or 1 1 p = 1 m; t = m; r = 1n; t = p 4. m = 1q; p = 1q; r = 1s; q = t; n = 1s a = 34 b = c = A = 9.5 sq units 84. A = 9.5 sq units (found using a calculator) A = 9.5 sq units

6 104 Applications of Trigonometry and Vectors Yes, vector addition is associative. 18. Yes, vector addition is commutative. 19. (a) 4,16 (b) 1,0 11. (c) 8, 8 0. (a) 4, 8 (b) 1,0 (c) 4, (a) 8, 0 (b) 0,16 (c) 4, 8. (a) 4,0 (b) 1, (c) 4, 4 3. (a) 0,1 (b) 16, 4 (c) 8, (a) 4, 4 (b) 1, 1 (c) 8, (a) 4i (b) 7i+ 3j (c) 5i+ j 6. (a) i+ 4j (b) i+ j (c) 4i 7j

7 Section 7.4: Vectors, Operators, and the Dot Product (a), 4 (b) 7, Magnitude: 16 ; Angle: 315 (θ lies in quadrant IV) 37. x 47; y 17 (c) 6, 6 8. (a) 4, (b) 13, x 17 y 0 x 38.8 y 8.0 (c) 3, 5 9. u = 1, w = 0, θ = x y x y u = 8, w = 1, θ = 0 4. x y u = 5 3 5, 31. u = 0, w = 30, θ = u = 4, v 3.064, v 1.984, u = 50, w = 70, θ = v , v 1.531, Magnitude: 17 ; Angle: (θ lies in quadrant IV) 34. Magnitude: 5 ; Angle: (θ lies in quadrant II) 35. Magnitude: 8 ; Angle: 10 (θ lies in quadrant II) newtons. (rounded to two significant digits) newtons. (rounded to two significant digits) lb. (rounded to three significant digits) lb. (rounded to three significant digits) 53. v 94. lb

8 106 Applications of Trigonometry and Vectors 54. v lb 55. v 4.4 lb 56. v lb 57. u+ v = a+ c, b+ d = ( + ) + ( + ) = ( + ) + ( + ) z z a bi c di a c b d i 1 Additional answers will vary. 59. u+ v =, u v = 6, 61. 4u = 8, v = 0, u 6v = 30, u+ 4v = 0, 65. u+ v 3u = 8, u+ v 6v = 4, , 8 = 5i+ 8j 68. 6, 3 = 6i 3j 69., 0 = i+ 0j= i 70. 0, 4 = 0i 4j= 4j θ = θ = θ = θ θ the vectors are orthogonal. 88. the vectors are not orthogonal. 89. the vectors are not orthogonal 90. the vectors are orthogonal. 91. the vectors are not orthogonal. 9. the vectors are not orthogonal 93. Draw a line parallel to the x-axis and the vector u + v (shown as a dashed line) Since θ 1 = 110, its supplementary angle is 70. Further, since θ = 60, the angle α is = 80. Then the angle CBA becomes 180 ( ) = = 30. Magnitude: The direction angle is ab, 4.104, cd,.509,.9544.

9 Chapter 7: Exercises , Magnitude: ; Angle: (θ lies in quadrant II) 98. They are the same. Preference of method is an individual choicesection 7.5: Applications of Vectors Chapter 7.5: Exercises lb at an angle of 167. with the 1480-lb force.. T 1800 lb at an angle of with the 840- lb force lb and the magnitude of the resultant is about 83 lb. 6. The magnitude of the resultant is 117 lb; the second force is 93.9 lb lb tons lb The weight of the crate is 64.8 lb; the tension is 61.9 lb. 14. (1) The angle between the 760-lb and 980-lb forces is 91.9 () The angle between the 10-lb and 980-lb forces is mi.; mi.; km km 19. The speed of the current is 3.5 mph and the actual speed of the motorboat is 19.7 mph. 0. The pilot turned at 1 hr 1 min after P.M., or at 3:1 P.M. 1. The bearing and ground speed of the plane. v The bearing is 37. (a) 6.3 (b) 14 seconds (c) 5 3. Let x = the airspeed and d = the ground speed. x 156 mph; d 161 mph The bearing is ; the ground speed is 0 mph. 6. The bearing is ; the ground speed is 181 mph. 7. The airspeed must be 170 mph. The bearing must be approximately The ground speed is 198 mph. Thus, the bearing is The ground speed is 30 km per hr. Thus, the bearing is (a) v t 56 mi sec. (b) 87 mi/sec. (3) The angle between the 760-lb and 10-lb forces is 16.6

10 108 Applications of Trigonometry and Vectors 31. (a) R = 5. and A = About. in. of rain fell. The area of the opening of the rain gauge is about 1.1 in.. (b) V = 1.5 ; The volume of rain was 1.5 (c) R and A should be parallel and point in opposite directions. a = a1, a, b = b1, b, and a b= a b, a b 1 1 a b = a + b a b cosθ a b = abcosθ Summary Exercises on Applications of Trigonometry and Vectors 3 in.. 1. The length of the two wires are about 9 ft and 38 ft.. TM 43 ft cm m apart. 5. x 15.8 ft per sec ; lb ft above the ground 8. (a) The speed of the wind is 10 mph (b) This represents a 30 mph wind in the direction of v. (c) u represents a southeast wind of 11.3 mph 9. Since 1 sin A 1, the triangle cannot exist. 10. Other angles can be ; third side 40.5, or other angles can be 143º50, 8º00 and third side 6.5.(Lengths in yards.) Chapter 7: Review Exercises m. B B = m ' or 15 40' 6. A = No; If you are given two angles of a triangle, then the third angle is known since the sum of the measures of the three angles is 180. Since you are also given one side, there will only be one triangle that will satisfy the conditions. 8. No; the sum of a and b do not exceed c. 9. a = 10, B = 30 (a) b = 10 sin 30 = 5. Also, any value of b greater than or equal to 10 would yield a unique value for A. (b) Any value of b between 5 and 10, would yield two possible values for A. (c) If b is less than 5, then no value for A is possible. 10. A = 140º, a = 5, and b =7 With these conditions, we can try to solve the triangle with the law of sines. sin A sin140 = = b a 7 5 7sin140 = B 64 5 Since A+ B = = 04 > 180, no such triangle exists. 11. A or ft m 14. B 6.5 or cm B = 17.3 ; C = ; c 11.0 yd 18. B 1 = 74.6 ; C 1 = 43.7 ; c m ; C = 1.9 ; c 0.0 m 19. c 18.65cm ; B 45 50' ; A = B 73.9 ; A 47.7 ; C = ,600 m ft.34 km 680 m feet feet long meters tall

11 Chapter 7: Review Exercises ft apart ft 30. C km 3..4 miles sq units 34. The sides of the triangle measure 5, 10, and 5 5, which satisfy the converse of the Pythagorean theorem. The area is 5 sq units 35. a b 36. a + 3c 48. The force is 80 newtons (rounded) at an angle of 30.4 with the first rope lb The pilot should fly on a bearing of 306. Her actual speed is 54 mph. 5. The resulting speed is 1 km per hr (rounded) and bearing is = AB ft and BC ft 54. Both expressions equal Both expressions equal 1 3 Chapter 7 Test km sq units sq units 37. (a) true (b) false newtons lb 40. horizontal: 5 ; vertical: horizontal: 869 ; vertical: magnitude: 9 Angle: (θ lies in quadrant IV) 43. magnitude: 15 ; Angle: 16.9 (θ lies in quadrant II) 44. (a) i (b) 4i j (c) 11i 7j 45. 9; , , (a) b > 10 (b) none (c) b a 40 m ; B 41 ; C = 79 B or B = Solving separately for triangles A 1 = a in. ; A = 0 00 ; a 431 in. 9. magnitude: v = 10 ; angle: 16.9 (θ lies in quadrant II) (a) 1, 3 (b) 6, 18 (c) 0 (d) 10

12 110 Applications of Trigonometry and Vectors 1..7 miles off the ground , mi m lb Chapter 7: Quantitative Reasoning 1 1. We can use the area formula A = sin rr B for this triangle. By the law of sines, we have r R Rsin A = r = sin A sin C sin C Since sin C = sin 180 ( A+ B) = sin ( A+ B), we sin sin have r = R A r = R A sin C sin ( A+ B) By substituting into our area formula, we have 1 A = rr sinb 1 Rsin A A = RsinB sin( A+ B) 1 sin AsinB A = R sin ( A+ B) Since there are a total of 10 stars, the total area covered by the stars is 1 sin A sin Asin B A = 10 R = 5 R sin( A+ B) sin( A+ B) R 3. (a) (b) 8.77 in. 5.3 in. (c) red